TITLE: Is $\mathbb{Q}$ homeomorphic to its open subintervals?
QUESTION [5 upvotes]: It is well known that $\mathbb{R}$ is homeomorphic to every interval $(a,b) \subset \mathbb{R}$. What can we say about $\mathbb{Q}$? Is it homeomorphic to some of its subintervals?

REPLY [2 votes]: Yes, and there are three ways that one can show that:

You can first prove that a countable linear order which is both dense and without endpoints is order isomorphic to the rational numbers. Therefore homeomorphic in the order topology to the rational numbers.
You can also prove Sierpinski's theorem that every countable metric space which has no isolated points is homeomorphic to the rational numbers. (See proof here, for example).'
You can find an explicit bijection. For bounded intervals you can use Thomas Andrew's suggestion from the comments, and for unbounded intervals MJD's suggestion reduces the case for a bounded interval again.